标签：des style blog http java color
You are given an N*N digit matrix and you can get several horizontal or vertical digit strings from any position.
For example:
123
456
789
In first row, you can get 6 digit strings totally, which are 1,2,3,12,23,123.
In first column, you can get 6 digit strings totally, which are 1,4,7,14,47,147.
We want to get all digit strings from each row and column, and write them on a paper. Now I wonder the sum of all number on the paper if we consider a digit string as a complete decimal number.
The first line contains an integer N. (1 <= N <= 1000)
In the next N lines each line contains a string with N digit.
Output the answer after module 1,000,000,007(1e9+7)。
3 123 456 789
2784
题目主要是导出公式：
如n行n列的每一行的和sum=1111.....111（n个1）*A1+111...111（n-1个1）*2*A2+.........+11*(n-1)*An-1+1*n*An;
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#define M 1000000007
#define f1(i, n) for(int i=1; i<=n; i++)
#define f2(i, n) for(int i=0; i<n; i++)
#define f3(j, n) for(int j=0; j<n; j++)
using namespace std;
char a[1050][1050];
long long int b[1050];
int main()
{
int n;
int k=1050;
f1(i, k)
b[i]=0;
f1(i, k)
for(int j=1; j<=i; j++)
b[i]=(b[i]*10+1)%M;
while(~scanf("%d",&n))
{
k=n;
long long int sum=0;
f2(i, k)
{
scanf("%s",&a[i]);
// f3(j, k)
for(int j=0; j<n; j++)
sum=(sum+((((j+1)*((long long)a[i][j]-'0'))*b[n-j])%M))%M;
}
//f3(j, k)
for(int j=0; j<n; j++)
{
f2(i, k)
sum=(sum+((((i+1)*((long long)a[i][j]-'0'))*b[n-i])%M))%M;
}
cout<<sum<<endl;
}
return 0;
}
标签：des style blog http java color
原文地址：http://blog.csdn.net/u013487051/article/details/37721409
